Solve the following system of linear equations using the matrix inverse method
3x + 2y = 7
4x + 3y =10
To solve the system of linear equations using the matrix inverse method, we need to represent the system in matrix notation.
Let's rewrite the equations in the form of AX = B, where A is the coefficient matrix, X is the variable matrix, and B is the constant matrix.
The given system of equations:
3x + 2y = 7 --> 3x + 2y - 7 = 0
4x + 3y = 10 --> 4x + 3y - 10 = 0
Now we can write the coefficient matrix A and the constant matrix B:
A = [3 2]
[4 3]
B = [7]
[10]
To find the variable matrix X, we can use the formula X = A^(-1) * B.
First, we need to find the inverse of matrix A. To do this, we calculate the determinant of A and check if it is nonzero. If the determinant is nonzero, A^(-1) exists.
The determinant of matrix A is calculated as follows:
det(A) = (3 * 3) - (2 * 4) = 1
Since the determinant is nonzero, A^(-1) exists. We can proceed to find the inverse of matrix A:
A^(-1) = 1/det(A) * adj(A)
Where adj(A) is the adjugate of matrix A, which is obtained by swapping the elements of the main diagonal and changing the sign of the other elements. For 2x2 matrices, the adjugate can be calculated as follows:
adj(A) = [d -b]
[-c a]
Where a, b, c, and d are the elements of matrix A.
For our matrix A, the adjugate is:
adj(A) = [3 -2]
[-4 3]
Now, we can calculate the inverse of matrix A:
A^(-1) = 1/det(A) * adj(A)
= (1/1) * [3 -2]
[-4 3]
= [3 -2]
[-4 3]
Now that we have the inverse of matrix A, we can find the variable matrix X:
X = A^(-1) * B
= [3 -2] * [7]
[10]
Multiplying the matrices, we get:
X = [3*7 + (-2)*10]
[-4*7 + 3*10]
X = [21 - 20]
[-28 + 30]
X = [1]
[2]
Therefore, the solution to the system of equations is x = 1 and y = 2.